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I've been trying to make a CFG, and npda/pda for this language (to construct an npda for accepting the language):

L(M)={ww:w∈{a,b}∗,|w| is even}.

i had already solved the reverse of the language (like this):

L′={wwR:w∈{a,b}∗},

Please check this snapshot to know what i am asking to do ! (L') : http://i921.photobucket.com/albums/ad53/Johann_1990/IMG_20150117_132616.jpg

but is there is a way to solve this L(M)={ww:w∈{a,b}∗,|w| is even}?

For example, abaaba∈L with w=aba.

Note that I'm not asking to prove by conduction or anything else.

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  • $\begingroup$ First of all, why should $|w|$ be even - it's not in your example. Then, $L'$ is a completely different language, and if I'm not totally wrong, $L$ is NOT context-free. So there is no pda/cfg. $\endgroup$ – lukas.coenig Jan 17 '15 at 12:55
  • $\begingroup$ I'm just giving an example to show what i am asking exactly, |w| is another example I'm trying to understand it too. I'm looking at the book page 178 "An introduction to formal languages and Automata 5th" its solving L as in my example ! $\endgroup$ – AaoIi Jan 17 '15 at 13:05
  • $\begingroup$ Can you post the exact text of the exercise (or whatever it is) and its context? Like this, it does not make sense. $\endgroup$ – lukas.coenig Jan 17 '15 at 17:25
  • $\begingroup$ @lukas.coenig the example context says : "to construct npda's that accept this language" : L(M)={ww:w∈{a,b}∗,|w| is even}. $\endgroup$ – AaoIi Jan 17 '15 at 18:55
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Short answer: $L$ is not context-free, so there is no CFG/PDA.

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  • $\begingroup$ You are right, i got it ! thank you :) $\endgroup$ – AaoIi Jan 17 '15 at 19:06
  • $\begingroup$ Nice, I'm happy for you! :-) $\endgroup$ – lukas.coenig Jan 17 '15 at 19:10

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